United States presidential election in Rhode Island, 1912
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The 1912 United States Presidential Election in Rhode Island took place on November 5, 1912 as part of the 1912 United States Presidential Election which was held throughout all contemporary 48 states. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.
Rhode Island was won by the Republican nominees, New Jersey Governor Woodrow Wilson and Indiana Governor Thomas R. Marshall. Wilson and Marshall defeated incumbent President William Howard Taft, and his running mate Vice President James S. Sherman and Progressive Party candidates, former President Theodore Roosevelt and his running mate California Governor Hiram Johnson.
Wilson won Rhode Island by a narrow margin of 3.48 percent, becoming the first Democratic presidential candidate since Franklin Pierce in 1852 to win the state.
Results
United States presidential election in Rhode Island, 1912[1] | ||||||||
---|---|---|---|---|---|---|---|---|
Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Democratic | Woodrow Wilson of New Jersey | Thomas Riley Marshall of Indiana | 30,412 | 39.04% | 5 | 100.00% | ||
Republican | William Howard Taft of Ohio | Nicholas Murray Butler of New York | 27,703 | 35.56% | 0 | 0.00% | ||
Progressive | Theodore Roosevelt of New York | Hiram Warren Johnson of California | 16,878 | 21.67% | 0 | 0.00% | ||
Socialist | Eugene Victor Debs of Indiana | Emil Seidel of Wisconsin | 2,049 | 2.63% | 0 | 0.00% | ||
Prohibition | Eugene Wilder Chafin of Illinois | Aaron Sherman Watkins of Ohio | 616 | 0.79% | 0 | 0.00% | ||
Socialist Labor | Arthur Elmer Reimer of Massachusetts | August Gillhaus of New York | 236 | 0.30% | 0 | 0.00% | ||
Total | 77,894 | 100.00% | 5 | 100.00% |
References
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